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Question

If xnπ2 and |cosx|sin2x3sinx+2=1 then all solutions of x are given by

A
2nπ+π2
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B
(2n+1)ππ2
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C
nπ+(1)nπ2
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D
none of these
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Solution

The correct option is B none of these
|cosx|sin2x3sinx+2=1
For equation to hold,
Case I:|cosx|=1
x=nπ,nI
Case II:|cosx|1
sin2x3sinx+2=0

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